Question

Two resistances are connected in parallel. Equivalent resistance of combination is \(\tfrac{6}{5}\ \Omega\). When one resistance is broken out, resistance becomes \(2\ \Omega\). The resistance of broken wire was:

• A. \(\tfrac{2}{5}\ \Omega\)
• B. \(2\ \Omega\)
• C. \(3\ \Omega\)
• D. \(\tfrac{5}{2}\ \Omega\)

Hint:

The remaining \(2\ \Omega\) is one resistor; put it into \(\frac{R_1R_2}{R_1+R_2}=\frac{6}{5}\) and solve.

Answer 1

The Correct Option is C

Step 1: Use conductances instead of resistances.
For parallel resistors, total conductance \(G = G_1 + G_2\) where \(G = 1/R\).
Step 2: Total conductance of the combination.
\(G = \dfrac{1}{6/5} = \dfrac{5}{6}\ \Omega^{-1}\).
Step 3: The intact resistor left after the break has \(R_1 = 2\ \Omega\), so \(G_1 = \dfrac{1}{2} = \dfrac{3}{6}\ \Omega^{-1}\).
Step 4: Conductance of the broken wire.
\(G_2 = G - G_1 = \dfrac{5}{6} - \dfrac{3}{6} = \dfrac{2}{6} = \dfrac{1}{3}\ \Omega^{-1}\).
Step 5: Invert to get its resistance.
\(R_2 = \dfrac{1}{G_2} = 3\ \Omega\).
\[\boxed{R_{\text{broken}} = 3\ \Omega}\]